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Home » The difference between the squares of two numbers is     . The square of the smaller number is     times the larger number. Determine the numbers.
The difference between the squares of two numbers is

    \[45\]

. The square of the smaller number is

    \[4\]

times the larger number. Determine the numbers.
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The difference between the squares of two numbers is

    \[45\]

. The square of the smaller number is

    \[4\]

times the larger number. Determine the numbers.

Let us consider the larger number be ‘x’

Smaller number be ‘y’

So according to the question,

    \[{{x}^{2}}~\text{ }{{y}^{2}}~=\text{ }45\]

… (i)

    \[{{y}^{2}}~=\text{ }4x\]

… (ii)

Now substitute the value of y in equation (i), we get

    \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }4x\text{ }=\text{ }45 \\ {{x}^{2}}~\text{ }4x\text{ }\text{ }45\text{ }=\text{ }0 \\ \end{array}\]

let us factorize,

    \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }9x\text{ }+\text{ }5x\text{ }\text{ }45\text{ }=\text{ }0 \\ x\left( x\text{ }\text{ }9 \right)\text{ }+\text{ }5\text{ }\left( x\text{ }\text{ }9 \right)\text{ }=\text{ }0 \\ \left( x\text{ }\text{ }9 \right)\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }=\text{ }0 \\ \end{array}\]

So,

    \[\begin{array}{*{35}{l}} \left( x\text{ }\text{ }9 \right)\text{ }=\text{ }0\text{ }or\text{ }\left( x\text{ }+\text{ }5 \right)\text{ }=\text{ }0 \\ x\text{ }=\text{ }9\text{ }or\text{ }x\text{ }=\text{ }-5 \\ \end{array}\]

When x =

    \[9\]

, then

The larger number = x =

    \[9\]

Smaller number = y =>

    \[{{y}^{2}}~=\text{ }4x\]

    \[y\text{ }=~\surd 4x\text{ }=~\surd 4\left( 9 \right)\text{ }=~\surd 36\text{ }=\text{ }6\]

When x =

    \[-5\]

, then

The larger number = x =

    \[-5\]

Smaller number = y =>

    \[{{y}^{2}}~=\text{ }4x\]

    \[y\text{ }=~\surd 4x\text{ }=~\surd 4\left( -5 \right)\text{ }=~\surd -20\]

(which is not possible)

∴ The value of x and y are

    \[9,6\]

.

i

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